A substance has a heat of fusion of 155 J/g. How many joules of energy are required to be removed in order to freeze a 10.0 g sa
1 answer:
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Complete Question
A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane
The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?
Answer:
The mass of caffeine extracted is
Explanation:
From the question above we are told that
The K value is
The mass of the caffeine is
The volume of water is
The volume of caffeine is
The number of times the extraction was done is n = 3
Generally the mass of caffeine that will be extracted is
substituting values
Answer:
(a) the mass of the water is 3704 g
(b) the mass of the water is 199, 285.7 g
Explanation:
Given;
Quantity of heat, H= 8.37 x 10⁶ J
Part (a) mass of water (as sweat) need to evaporate to cool that person off
Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg
H = m x Lvap.
mass in gram ⇒ 3.704 kg x 1000g = 3704 g
Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J
specific heat capacity of water, C, 4200 J/kg.°C
H = mcΔθ
where;
Δθ is the change in temperature = 35 - 25 = 10°C
mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g